2463. Minimum Total Distance Traveled - Detailed Explanation

Brute Force – Java Implementation:

Complexity Analysis of Brute Force: In the worst case, the brute-force approach is exponential in time. Each robot has to consider assignments to potentially each factory, and each factory can take multiple robots, leading to an enormous search tree. Even with pruning, the time complexity can be approximated as O(\text{exponential in } n+m), which is infeasible for n,m up to 100. For example, if we simplified the problem to each of n robots choosing among m factories, that’s m^n possibilities. The backtracking approach prunes some impossible combinations (due to capacity limits and sorted order), but it is still far too slow for large inputs. The space complexity is O(n+m) for the recursion stack (in the worst case we might recurse depth n+m) and additional space for storing the input and copies. The brute force doesn’t use significant extra memory aside from recursion, but it’s impractical due to time complexity.

Optimized Approach (Dynamic Programming)

Idea: We can optimize by recognizing overlapping subproblems and using dynamic programming. The key observation from the sorted order and contiguous assignment hint is that if both robots and factories are sorted by position, we will assign robots to factories in order. This means each factory will take a consecutive block of robots from the sorted list. We can build the solution factory by factory. Define a DP state dp[i][j] = the minimum total distance to repair the first i robots using the first j factories (where i and j range from 0 up to n and m respectively) . We want to compute dp[n][m], the cost for all n robots with all m factories.

DP Transition:
Consider the j-th factory (at position p with capacity L). This factory can repair anywhere from 0 up to L robots, but it cannot exceed i (the total robots considered) either. Moreover, because we assign in order, if this j-th factory repairs k robots, it will be the last k robots out of the first i (since earlier factories would have handled the others). For example, if we are assigning the first 10 robots among the first 3 factories, and factory 3 fixes k=4 of them, it would take robots number 7,8,9,10 (assuming robots are numbered in sorted order).

Therefore, we have a recurrence:

• Let factory j fix k robots (where 0 \le k \le \min(L, i)). These k would be robots indexed i-k+1 through i in the sorted order. The total distance for these k robots to go to factory j is the sum of distances from each of those robot positions to p (the factory’s position). Call that sum dist(k, j) for k robots assigned to factory $j`.

• If factory j takes these k robots, then the first i-k robots must have been handled by the first j-1 factories. The cost of that scenario would be dp[i-k][j-1] + dist(k, j).

• We choose the k that gives the minimum cost. We also consider k=0 (meaning factory j fixes no one and we rely on previous factories to cover i robots). This gives the transition:

[dp[i][j] = \min_{0 \le k \le \min(i,\, limit_j)} \ Big( dp[i-k][j-1] + \text{sum of distances for the last k of these i robots to factory j} \Big). ]

We fill this DP table iteratively. Base cases: dp[0][j] = 0 for all j (0 robots requires 0 distance regardless of factories), and dp[i][0] = ∞ (if no factories are available but i>0 robots need repairs, that’s impossible). We iterate over factories and robot counts to compute minimal distances.

To efficiently compute the distance sum for a given k robots to a factory: since robots and factories are sorted by position, the k robots that factory j would fix are contiguous in sorted order. We can pre-sort the robot array and use prefix sums to get the sum of positions quickly. However, because the distance is absolute difference from the factory’s position, we actually compute it on the fly in the DP loop for simplicity (since n is at most 100, this is manageable).

Alternatively, one could precompute a helper table cost[j][k] = the total distance if factory j fixes k robots starting from some index, but it’s not necessary unless we want to micro-optimize.

Dynamic Programming – Python Implementation:

Explanation of the DP approach: We build up solutions for increasing numbers of robots and factories. For each factory, we consider how many robots it will serve. By iterating possible k (robots assigned to that factory) and looking up a previously computed subproblem for the remaining robots (dp[i-k][j-1]), we efficiently explore combinations without redundant recursion. Sorting ensures that we always assign the closest available robots to each factory segment, because assigning non-consecutive robots to one factory would imply some closer robot was handled by a farther factory, which would only increase total distance.

For example, in Example 2 (robots at -1 and 1, factories at -2 and 2): after sorting we have robots [-1, 1], factory1 at -2 (limit1), factory2 at 2 (limit1). Our DP starts with dp[0][*] = 0. For the first factory (at -2):

• dp[1][1] (1 robot with factory1) = min of either factory1 fixes 0 robots (then 1 robot with 0 factories = impossible) or fixes 1 robot. If it fixes 1, cost = distance from robot -1 to -2 = 1. So dp[1][1] = 1.
• dp[2][1] (2 robots with only factory1) = factory1 can fix at most 1, so it cannot handle 2 robots alone (we get infinity for that state).

Now include factory2 (at 2):

• dp[1][2] (1 robot with both factories) = min(dp[1][1] (which is 1), or have factory2 fix that one robot with cost = 3). This yields dp[1][2] = 1 (it’s better for factory1 to have handled the single robot).
• dp[2][2] (2 robots with both factories) = consider factory2 fixing 0 or 1 robot. If factory2 fixes 0, cost = dp[2][1] which was ∞ (not possible). If factory2 fixes 1, then factory1 must have fixed 1 (the other robot): cost = dp[1][1] + dist where dp[1][1]=1 and dist = distance from the remaining robot (which would be the second robot at 1) to factory2 at 2 = 1. So total = 1+1 = 2. Thus dp[2][2] = 2. This matches the answer. We efficiently found that distribution (each factory fixes one robot) is optimal.

In Example 1 (robots at 0,4,6 and factories at 2(cap2), 6(cap2)): the DP would find after the first factory that dp[2][1] = 4 (factory at 2 fixing two robots 0 and 4) and dp[1][1] = 2 (factory at 2 fixing one robot 0) are the viable states (3 robots with one factory was impossible). Then including the second factory, it would compute dp[3][2] by considering the second factory fixing 1 robot or 2 robots. Either scenario gives 4 total distance in this case, which is the optimal result. The DP table approach systematically evaluates these possibilities.

Time Complexity: The DP solution runs in polynomial time. We fill an n+1 by m+1 table and for each cell potentially iterate up to limit_j times. In the worst case, a factory’s limit could be n (e.g., a factory can repair all robots), so the innermost loop runs up to n. This gives a complexity on the order of O(n * m * n) = O(n^2 * m). Since n, m <= 100`, this is at most on the order of 100^2 * 100 = 1,000,000operations, which is very efficient. We can also describe this asO(n * m * L)whereLis the maximum limit value. For our constraints, that’sO(100 * 100 * 100) = 10^6$.

Space Complexity: We used a 2D DP array of size (n+1) \times (m+1), which is 101 \times 101 \approx 10^4 elements in the worst case. This is O(n * m) space, which is negligible. We can optimize space further by noticing that to compute dp[*][j] we only need the previous column dp[*][j-1]. Thus, we could use two 1D arrays of length n+1 (or even one array updated in place in reverse order) to reduce space to O(n). However, given the small size, the 2D table is fine and easier to understand.

Step-by-Step Walkthrough

Let’s walk through a small example to see how the solution works step by step. We’ll use Example 2 for clarity (two robots and two factories):

• Step 1: Sort inputs.
  Robots: [-1, 1] (sorted order)
  Factories: [(-2, limit=1), (2, limit=1)] (sorted by position)

• Step 2: Initialize DP table.
  We create a table dp[i][j] where i ranges from 0 to 2 (number of robots) and j from 0 to 2 (number of factories). Initially, dp[0][j] = 0 for all j` (0 robots requires 0 distance), and `dp[i][0] = ∞` for i>0$ (no factory cannot fix robots). So starting state (some entries omitted for brevity):

  dp[0][0] = 0   dp[0][1] = 0    dp[0][2] = 0  
  dp[1][0] = ∞   dp[1][1] = ∞    dp[1][2] = ∞  
  dp[2][0] = ∞   dp[2][1] = ∞    dp[2][2] = ∞  
  

• Step 3: Consider Factory 1 (position -2, limit 1). We fill column for j=1 (using only the first factory):

  • For 0 robots: dp[0][1] = 0 (base case).
  • For 1 robot: dp[1][1] – Factory 1 can fix at most 1 robot. We have two options: fix 0 robots or 1 robot.
    • If it fixes 0 robots, we look at dp[1][0] which is ∞ (impossible).
    • If it fixes 1 robot, it will fix the 1st robot (at -1). Distance = |–1 – (–2)| = 1. Add that to dp[0][0] (cost for remaining 0 robots with 0 factories) = 0 + 1 = 1.
    • Take min of options ⇒ dp[1][1] = 1.
  • For 2 robots: dp[2][1] – Factory 1 cannot fix 2 because its limit is 1. The options:
    • 0 robots by factory1 ⇒ check dp[2][0] = ∞.
    • 1 robot by factory1 ⇒ then 1 robot is left for 0 factories (dp[1][0] = ∞). Even though distance for 1 robot would be 1, the subproblem dp[1][0] is impossible, so this route is invalid.
    • Thus, dp[2][1] remains ∞ (it’s not possible to repair 2 robots with only the first factory).
      After factory 1, the DP table (relevant entries) is:

  dp[1][1] = 1  
  dp[2][1] = ∞  
  

• Step 4: Consider Factory 2 (position 2, limit 1). Now include the second factory (fill column for j=2):

  • For 1 robot: dp[1][2]. We have two factories for one robot. Options:

    • If factory2 fixes 0 robots, then dp[1][2] = dp[1][1] = 1 (the cost is already covered by factory1).
    • If factory2 fixes 1 robot, then factory1 fixes 0 (look at dp[0][1]) and distance = |1 – 2| = 1. dp[0][1] is 0, so this option cost = 0 + 1 = 1.
    • Both options yield 1. Min ⇒ dp[1][2] = 1. (It basically means either factory could have handled that single robot at total cost 1; the algorithm will naturally pick the minimum which is 1 either way.)

  • For 2 robots: dp[2][2]. Options:

    • Factory2 fixes 0 robots ⇒ then both robots must be handled by factory1: cost = dp[2][1] which we found was ∞ (not feasible).
    • Factory2 fixes 1 robot ⇒ then factory1 fixes 1 robot (since total 2 robots, one handled by factory2, the other by factory1). We take the scenario of factory1 handling 1 (which we have as dp[1][1] = 1) and factory2 handling the other. Which robot will factory2 handle? Since we assign in order, if each factory handles one, factory1 will take the first robot (-1) and factory2 will take the second robot (1). Distance for the robot at 1 to go to factory2 at 2 is 1. So this option cost = dp[1][1] + 1 = 1 + 1 = 2.
    • Factory2 fixes 2 robots (its limit is 1, so this is not allowed).
    • Minimum of the valid options ⇒ dp[2][2] = 2.

Now dp[2][2] is computed as 2, which is our answer. The DP table effectively tried all meaningful assignments (first factory took one robot and second took the other in this case) in a structured way. We avoided brute-forcing all combinations by using the results of subproblems (like reusing the result that the first factory handling 1 robot gives cost 1). The process for larger inputs is similar: we add factories one by one and distribute the robots among them optimally at each step.

Common Mistakes & Edge Cases

• Greedy assignment pitfall: A common mistake is to try a greedy strategy such as always sending each robot to the nearest available factory. This can fail because using a nearby factory for one robot might prevent that factory from fixing several other robots that are even closer. For example, suppose one factory has a small limit but many nearby robots, and another factory is slightly farther but has more capacity – a greedy choice might fill the small factory with a robot that the larger factory could have handled, and then other robots end up traveling much farther. The optimal solution requires considering the distribution globally, which is why DP is needed. Always check counterexamples before assuming a greedy approach works.

• Not sorting the inputs: If you don’t sort the robots and factories by position, it’s hard to ensure contiguity of assignments. The DP formulation (or any logical assignment strategy) relies on sorted order. Without sorting, you might double-count or assign out-of-order which complicates the problem. Ensure to sort both arrays first, otherwise the logic of “first i robots” or “first j factories” in DP won’t correspond to the smallest-distance pairing.

• Misusing dynamic programming indices: It’s easy to mix up indices when implementing the DP. Remember that dp[i][j] refers to count of robots and factories, not direct array indices. If using 0-based indexing for arrays, be careful when computing distances for the “last k robots” – you might need to offset indices correctly. Off-by-one errors can lead to incorrect cost calculations or index out of range. A good practice is to clearly document what each dimension of DP represents (e.g., i = number of robots considered, so those robots are robot[0]...robot[i-1] if using 0-indexed list).

• Resetting state between factories: In the recursion or DP, when you move to consider the next factory, ensure that the count of robots assigned to the current factory resets appropriately. For example, in a recursive solution, if you carry a variable for how many robots have been assigned so far to the current factory, make sure to reset it (or use separate states) when moving to the next factory. The DP approach we used avoids this by integrating the count into the state. But if one implemented a memoization with a state (i, j, k) (robot index, factory index, how many already assigned to current factory), forgetting to reset k when advancing to the next factory would be a bug.

• Handling impossible states: Make sure that impossible states remain marked as “infinite” (or a very large number). For instance, dp[i][0] for any i > 0 should stay as infinity (or a sentinel large value) to indicate it’s not achievable. And in code, when adding distances, be careful not to overflow if you’re using a large sentinel value (we used Long.MAX_VALUE/4 in Java to avoid overflow when adding distances).

• Edge cases: Consider edge scenarios to ensure your solution covers them:

  • All robots already at factory locations (distance should be 0 because no one needs to move).

  • A factory with limit = 0 (it can be completely ignored in the assignment because it can’t repair any robot).

  • Only one factory available (then that factory must fix all robots, so the answer is the sum of distances of all robots to that factory’s position – or infinity if it doesn’t have enough capacity, but the problem guarantees a valid assignment exists).

  • Only one robot (the answer will just be the distance from that robot to the nearest factory that has capacity for it – the DP will handle this, but it’s good to mentally check if code handles single-element arrays correctly).

  • Factories with extremely large or small coordinates (to test that using abs(position difference) works within the data type limits and you didn’t, for example, square a number by mistake or cause overflow).

Alternative Variations & Related Problems

• Min-Cost Matching / Assignment Problem: This problem can be viewed as a form of the assignment problem where we want to assign n robots to m factory “slots” (each factory has multiple slots equal to its limit) with minimum cost. A more general solution could use a min-cost max-flow algorithm or the Hungarian algorithm for assignment if we construct a bipartite graph between robots and factory slots (each slot being an opportunity for a robot to be fixed at that factory). Our DP solution leverages the one-dimensional sorted structure for efficiency. In a more general metric space (say the robots and factories were in a 2D grid with Manhattan distances), a state-space search or min-cost flow would be a typical approach. For one-dimensional positions, the DP is simpler and faster.

• Variation – minimizing maximum distance: A different but related question could be: “If we wanted to minimize the maximum distance any robot travels (instead of the sum of distances), how would we approach it?” This would turn into a different problem (more like a radius or coverage problem) that might be solved with binary search on distance and a greedy check (or DP) for feasibility. It’s a variation to be aware of: sum and max optimizations often require different strategies.

• Similar LeetCode problems for practice:

  • Campus Bikes II (LeetCode 1066) – You have n workers and m bikes on a 2D grid and want to assign each worker a bike with minimum total Manhattan distance. This is similar in that it’s an assignment minimization problem, but each bike can only be used by one worker (capacity 1). The solution uses bitmask DP to handle the assignment since n is relatively small (or a Hungarian algorithm for the general case).

  • Candy Distribution (not a direct LeetCode problem, but a classic greedy problem) – It involves assigning candies to children with certain preferences, which is about matching resources to agents, albeit usually solved greedily for maximizing satisfaction. It’s conceptually related in terms of assigning units to agents.

  • Minimum Cost to Connect Two Groups of Points (LeetCode 1595) – This is another DP problem where we connect points in two sets with minimum cost, and each point in one group must be connected to at least one point in the other group. It can be thought of as an assignment with possible multiple connections. The state compression techniques there (bitmask DP) are somewhat analogous to assignment problems.

  • Assigning Tasks – There are various task assignment problems (like assigning jobs to machines or people) on LeetCode (e.g., “Task Scheduler”, “Two City Scheduling”). For instance, Two City Scheduling (LeetCode 1029) involves assigning people to two cities with costs, which can be solved by greedy or DP. While the context differs (minimizing travel cost to two cities vs. robots to factories), the pattern of distributing a set of entities between different “destinations” under constraints is similar.

  • If you are interested in the algorithmic concept, studying the Hungarian Algorithm for the assignment problem or Min-Cost Max-Flow for flow with costs will give you a more general toolkit to tackle variations of this problem where greedy or simple DP by sorting might not apply. The problem we solved is essentially a flow where each factory provides capacity (supply) and each robot is a demand of 1, with cost = distance; our DP is a specialized way to solve that due to the sorted line structure.